We can now consider that the number of flips to get 160 head

We can now consider that the number of flips to get 160 heads is equal to the number of flips it takes to get the first head, plus the number it takes to get the second, plus the number it takes to get the third... plus the number it takes to get the 160 h head. Since flips are independent, we have the result being the sum of 160 independent, identically distributed random variables, each representing the number of flips it takes to get just one head. What is the distribution which applies to the random variables being added up in the previous paragraph? What is the expected value of each of them? What is the variance of each of them? What is the standard deviation of each of them? Now since we have a sum of 160 random variables, it appears that the Central Limit Theorem should apply. Use the C\'LT (with continuity collection) to get an approximate numerical answer for the probability that it will take more than 295 flips. (The continuity correction is discussed in Section 7.2 of your text.) Your approximate answer should be reasonably close to the exact answer given above.

Solution

Let X be the number of flips to get 160 heads.

x1 = number of flips it takes to get first head.

x2 = number of flips it takes to get second head.

x3 = number of flips it takes to get third head.

.

.

.

x160 = number of flips it takes to get 160th head.

And given that,

X = x1 + x2 + x3 + .........+ x160

We know that the addtive property of binomial distribution.

X follows binomial distribution with parameters n = 160 an p = 1/2.

mean = p = 0.5

variance = [(p*q)/n] = (0.5*0.5)/160 = 0.001563

sd = sqrt(variance) = sqrt (0.001563) = 0.0395

n = number of heads = 160

By using CLT we have to find probability that it willl take more than 295 flips.

sample proportion (p^) = x / n = 160 / 295 = 0.5424

Now we have to find P(X > 0.5424)

Now we have to convert x = 0.5424 into standard normal distribution.

z = (x - mean) / sd

= (0.5424 - 0.5) / 0.0395

z = 1.0727

Now we have to find P(Z > 1.0727) = 1 - P(Z < = 1.0727)

This probability we can find by using EXCEL.

syntax :

NORMSDIST(z)

where z is the test statistic value.

P(Z <= 1.0727) = 0.8583

P(Z > 1.0727) = 1 - 0.8583 = 0.1417